2 group problems *TRICKY*
ron
group theory which I couldn't solve and
I think they're pretty tricky. Maybe some of
u will have better luck.
1. G is finite group not divisble by 3 and
for all a,b in G, (ab)^3 = (a^3)(b^3). Prove
G is abelian.
2. In any group G, the number of solutions to x^n = 1
(n divides size of G) is a multiple of n. I proved
this easily in abelian case, but it's kindof tough
in general.
Thanks for your help.
-r.j.
Arturo Magidin
>HI all, I have some problems from
>group theory which I couldn't solve and
>I think they're pretty tricky. Maybe some of
>u will have better luck.
>
>1. G is finite group not divisble by 3 and
>for all a,b in G, (ab)^3 = (a^3)(b^3). Prove
>G is abelian.
I assume you mean that the order of G is not divisible by 3. Note that
this means that the group is 3-divisible (i.e., every element has a
cubic root).
From (ab)^3=a^3b^3 you get that
(ab)^2 = a^3*b^3*(ab)^{-1} = a^3 b^2 a^{-1}
Therefore, abab = a^3 b^2 a^{-1}, so bab = a^2 b^2 a^{-1}, and we have
that for any a and b,
baba=(ba)^2 = a^2b^2.
We also know that (ba)^3 = b^3 a^3, and (ba)^3 = ba(ba)^2=ba(a^2b^2) =
b a^3 b^2.
So ba^3 b^2=b^3a^3, and that means that for any pair of elements a and
b,
a^3b^2 = b^2a^3.
So every cube commutes with every sqaure. However, as we noted above,
EVERY element of G is a cube, so every square in G is central in G.
Therefore, since (ab)^2 = a^3b^2a^{-1} = a^3 a^{-1}b^2 (since b^2 is
central), you get that (ab)^2 = a^2 b^2, which is a well known
sufficient condition for <a,b> to be abelian. This proves that G is
abelian.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
Arturo Magidin <mag...@math.berkeley.edu> wrote:
Sorry for following up on my own post...
>In article <ugbtsa...@forum.mathforum.com>, ron <rk...@yahoo.com> wrote:
>>HI all, I have some problems from
>>group theory which I couldn't solve and
>>I think they're pretty tricky. Maybe some of
>>u will have better luck.
>>
>>1. G is finite group not divisble by 3 and
>>for all a,b in G, (ab)^3 = (a^3)(b^3). Prove
>>G is abelian.
>
>I assume you mean that the order of G is not divisible by 3.
>Note that this means that the group is 3-divisible
>(i.e., every element has a cubic root).
I guess the other possible reading is the opposite: that G is finite,
but it is NOT 3-divisible. But in that case, the result would be
false: consider the split meta-cyclic group of order 27, given by
G = < x,y | x^3 = y^9 = e; yx=x(y^4) >.
It is not hard to verify that in this case, (x^i y^j)^3 = y^{3j}, and
so that for any a,b\in G, (ab)^3=a^3b^3; the group is not 3-divisible,
and the group is not abelian, since [y,x]=y^3 <> e.
ma...@mimosa.csv.warwick.ac.uk
rk...@yahoo.com (ron) writes:
>HI all, I have some problems from
>group theory which I couldn't solve and
>I think they're pretty tricky. Maybe some of
>u will have better luck.
>
>1. G is finite group not divisble by 3 and
>for all a,b in G, (ab)^3 = (a^3)(b^3). Prove
>G is abelian.
Arturo Magidin gave a slick solution to this one. He showed first that
(ab)^3 = a^3b^3 for all a,b in G implies that g^2h^3 = h^3g^2 for all
g,h, in G. The proof of that did not use the fact that |G| is not divisible
by 3. Here is a follow-up problem:
Let G be a torsion-free group in which (ab)^3 = (a^3)(b^3) for all a,b in G.
Prove that G is abelian.
>2. In any group G, the number of solutions to x^n = 1
>(n divides size of G) is a multiple of n. I proved
>this easily in abelian case, but it's kindof tough
>in general.
This is a theorem of Frobenius. You can find the proof in Marshall Hall's
"The Theory of Groups", Theorem 9.1.2.
Hall mentions a related conjecture:
Let G be a finite group, let n divide |G|, and suppose that the number of
solutions of x^n=1 in G is exactly n. Then the solution set forms a normal
subgroup of G.
Does anyone know whether this was ever settled? I have a vague memory
that somebody eventually proved it using the classification of finite
simple groups.
Derek Holt.
ron
Thanks for the help. There's only one problem, this exercise
is taken out of Herstein, and it's given after the first few
sections on groups. Nothing is assumed except basics such as
Lagranges theorem, and size of HG (H G subgroups). However,
his book is notorious for giving many impossible exercises.
Is there any other solution, or at least, can you point out
how to prove that if group's order isn't divisble by 3, then
every element is a cube? Thanks.
Btw...do u have any idea how to solve the other problem?
Arturo Magidin
>Hi Arturo,
>
>Thanks for the help. There's only one problem, this exercise
>is taken out of Herstein, and it's given after the first few
>sections on groups. Nothing is assumed except basics such as
>Lagranges theorem, and size of HG (H G subgroups). However,
>his book is notorious for giving many impossible exercises.
>
>Is there any other solution, or at least, can you point out
>how to prove that if group's order isn't divisble by 3, then
>every element is a cube? Thanks.
The following uses Lagrange's Theorem:
Lemma. Let G be a group of order n. For any k, if gcd(k,n)=1, then
every element of G is a $k$-th power.
Proof. Let x in G. Then <x> is a cyclic subgroup of G, and therefore
has order N, with N|n; in particular, gcd(N,k)=1.
Let a and b be integers such that aN+bk=1. Then
x = x^{aN+bk} = x^{aN}x^{bk} =(x^b)^k
therefore, x is the k-th power of x^b, and in particular it is a
$k$-th power. QED
Since 3 is prime, the fact that 3 does not divide the order of group
means that it is relatively prime to it, yielding the result.
>Btw...do u have any idea how to solve the other problem?
Prof. Holt has kindly given you a reference for it.
Arturo Magidin
<ma...@mimosa.csv.warwick.ac.uk> wrote:
[.snip.]
>Arturo Magidin gave a slick solution to this one. He showed first that
>(ab)^3 = a^3b^3 for all a,b in G implies that g^2h^3 = h^3g^2 for all
>g,h, in G. The proof of that did not use the fact that |G| is not divisible
>by 3. Here is a follow-up problem:
>
>Let G be a torsion-free group in which (ab)^3 = (a^3)(b^3) for all a,b in G.
>Prove that G is abelian.
This was actually pretty neat. Thanks for proposing it:
Consider the commutator of x and y in G, [x,y]=x^{-1}y^{-1}xy. Then,
in G, we have:
[x,y]^3 = x^{-3}y^{-3}x^3 y^3
= x^{-3}y^{-3}y^2 x^3 y (since squares commute with cubes)
= x^{-3}y^{-1}x^3 y
= [x^3,y]
for all x and y in G. Therefore, replacing y with an arbitrary square,
we have
[x,h^2]^3 = [x^3,h^2] = e (since cubes and squares commute).
Since G is torsionfree, we must have [x,h^2]=e, so again we conclude
that every square is central in G.
Now, we have for any a,b in G, that
(ab)^3 = a^3b^3 = a b^2 a^2 b (since squares are central)
so (ba)^2 = b^2 a^2, and we may again conclude that G is abelian from
this.
ron
Thank you for suggesting a reference...that problem has
been buggin me for a while now :)
Regarding your follow-up problem: does that require some
advanced theorems or very clever tricks, because I'm not
good at those. Maybe I can get this if you gave me a
hint. Thanks.
ma...@mimosa.csv.warwick.ac.uk
mag...@math.berkeley.edu (Arturo Magidin) writes:
>In article <9d8bic$49n$1...@wisteria.csv.warwick.ac.uk>,
> <ma...@mimosa.csv.warwick.ac.uk> wrote:
>
> [.snip.]
>
>>Arturo Magidin gave a slick solution to this one. He showed first that
>>(ab)^3 = a^3b^3 for all a,b in G implies that g^2h^3 = h^3g^2 for all
>>g,h, in G. The proof of that did not use the fact that |G| is not divisible
>>by 3. Here is a follow-up problem:
>>
>>Let G be a torsion-free group in which (ab)^3 = (a^3)(b^3) for all a,b in G.
>>Prove that G is abelian.
>
>This was actually pretty neat. Thanks for proposing it:
That's OK! Your solution is better than mine. Since squares commute with
cubes, it is easy to show that all sixth powers are central. I was then
going to use the fact that finitely generated groups of exponent 6 are
finite, but that is unnecessary now.
Derek Holt.
ron
G is finite, and (ab)^p = (a^p)(b^p) [p is prime dividing |G|]
show that Sylow-p subgroup is normal.
ma...@mimosa.csv.warwick.ac.uk
You don't need to assume that p divides |G|, because the result is trivial
if it does not !
Call an element a p-power element if its order is a power of p, and
a p'-element if its order is coprime to p.
We claim that the product of two p-power elements a,b of G is itself
a p-power element.
We do this by induction on o(a) - it is clearly true when o(a)=1.
If o(a) > 1, then (ab)^p = a^pb^p is a p-power element by inductive
hypothesis, but then ab is also a p-power element.
From the claim it follows that the set of all p-power elements form a
subgroup of G, which is easily seen to be a normal Sylow-p subgroup P.
In fact we can prove the stronger result that P is a direct factor of G.
By the same argument as in the p=3 case, we get
a^(p-1)b^(p-1) = (ba)^(p-1) => ba^pb^(p-1) = (ba)^p = b^pa^p =>
a^pb^(p-1) = b^(p-1)a^p i.e.
All p-th powers commute with all (p-1)-th powers.
(That statement is true even if p is not prime.)
Since p and (p-1) are coprime, each p-power element is a (p-1)-th power
and each p'-element is a p-th power, so p-power elements commute with
p'-elements.
By the Schur-Zassenhaus Theorem, P has a complement C in G,
and then C consists of p'-elements, so everything in C commutes with
everything in P, and we get G = C X P.
Derek Holt.